Valence Bond Theory, Effective Atomic Number & Magnetic Properties - Free

Due to strong ligand, this is inner orbital complex and the hybridisation of Fe is d²sp³.

Since, F^- ion is a weak ligand hence, pairing of electrons does not occur and outer orbital complex is formed in [CoF₆]^3-.

[Ni(CN)₄]^2- in this, Ni has dsp² hybridisation with square planar geometry and it is diamagnetic.

A magnetic moment of 1.73 BM corresponds to one unpaired electron. Among the given, (a) and (b) are diamagnetic and (c) is paramagnetic with 3 unpaired electrons (3d⁷ configuration).

F^- ion is weak ligand hence, pairing of electrons does not occur and outer orbital complex is formed in [CoF₆]^3-.

The potassium ferrocyanide, K₄ [Fe(CN)₆] due to strong ligand pairing of electrons occur and it has no unpaired electrons.

 In [Co(NH₃)₆]³⁺ due to pairing cobalt configuration and it is diamagnetic.
In [CoF₆]^3- , F^- being weak ligand pairing does not occur and Co³⁺ ion has configuration . It has 4 unpaired electrons.

ln [NiCI₄]^2- , Ni has sp³ hybridisation with 2 unpaired electrons, i.e. paramagnetic.
In [PdCI₄]^2- Pd has dsp² hybridisation and it has no unpaired electrons, i.e. diamagnetic.

Zn and Ni always give outer orbital complexes with coordination number 6.
[Co(NH₃)₆]³⁺ is diamagnetic complex.
[Cr(NH₃)₆]³⁺ is param agnetic with 3d³ configuration.

Correct option: D

Cu in [Cu(NH₃)₄]²⁺ is in the +2 oxidation state and has a d⁹ electronic configuration; for four-coordinate Cu(II) with ammine ligands a square-planar arrangement is commonly adopted, described by dsp² hybridisation.

Ni in [Ni(NH₃)₄]²⁺ is Ni(II) with a d⁸ configuration; four-coordinate d⁸ complexes typically form a square-planar geometry with dsp² hybridisation. Hence both complexes in option D have the same hybridisation: dsp².

Briefly rule out other options:

Option A: [Co(NH₃)₆]³⁺ is a low-spin d⁶ complex (inner-orbital) giving d²sp³, whereas [Co(H₂O)₆]³⁺ is typically high-spin d⁶ and described by outer-orbital sp³d²; they differ.

Option B: [NiCl₄]^2- is usually tetrahedral (sp³) while [PtCl₄]^2- is square planar (dsp²); they differ.

Option C: [Zn(NH₃)₄]²⁺ (Zn²⁺, d¹⁰) commonly adopts a tetrahedral sp³ arrangement, while [Cu(NH₃)₄]²⁺ is square planar (dsp²); they differ.

Therefore, option D is the only pair with the same hybridisation (dsp²).

The complex which involves d-orbitals of outer shell are called high spin complex as well as outer orbital complex.

[ZnCI₄]^2-, [CuCI₄]^2- and [Ni(CO)₄] have sp³ hybridisation and all three contains weak field ligands.

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K₂[Hgl₄] In this EAN = 80 - 2 + 8 = 86 (Equal to noble gas, radon).
[Pd(NH₃)CI₂] In this EAN = 46 - 2 + 8 = 52 (Not equal to noble gas, xenon)
[Cdl₄]^2- In this EAN = 48 - 2 + 8 = 54 (Equal to noble gas, xenon)
Co₂(CO)₈ In this EAN = Electrons from Co-atom (27) + electrons from 4 CO molecules (8) + one electron from Co— Co bond (1) = 36 (Equal to noble gas krypton)

[Fe(H₂O)₆]²⁺ and [Fe(CN)₆]^4- have different hybridisation, magnetic moment and colour also.

Cu(Z = 29) copper configuration (3d¹⁰4s¹)

Hence, [Cu(NH₃)₄]²⁺ has square planar geometry and 1 unpaired electron in Ad orbital which is transferred from 3d orbital with magnetic moment 1.73 BM.

It is inner orbital complex, diamagnetic.

(i) → (p.s), (ii) → (p,s), (iii) → (q,t), (iv) → (q.r)

A → (iii) (b) B → (i) (c) C → (iv) (d) D → (ii)

a) Six empty orbitals (two d, one s and three p) are available, so whether the ligand is weak (H2​O)  the six orbitals hybridize to give six equivalent d2sp3 hybrid orbitals. Hence, in  [Cr(H2​O)6​]3+  ions chromium is in a state of d2sp3 hybridization.

b)  CN is a strong ligand it makes the unpaired electrons of cobalt to pair up and occupies the space.

• The ligands occupy one d orbital, one s orbital and 2 p orbitals. Thus the hybridisation is “dsp2” with “square planar” geometry.
• It has 1 unpaired electron.

c) In [Ni(NH3)6]2+, Ni is in +2 state and has configuration 3d8. In presence of NH3, the 3d electrons do not pair up. The hybridization is sp3d2 forming an outer orbital complex.

 It has been found that the complex has two unpaired electrons.

Diamagnetic are
[Zn(OH)₄]^2-, K₄[Fe(CN)₆], [PdBr₄]^2-,[Ni(CO)₄]

In this, Mn has +3 oxidation state. Mn³⁺ configuration = [Ar]3d⁴. This forms outer complex. Hence, unpaired electrons are 4.

R[44] = [Kr]4d⁷ 5s¹ (in ground state)
In Ru[2+] => 4d⁶ 
=> (t₂g)⁶ (eg)⁰

^{Hence, no unpaired electron. Therefore, correct answer is zero.}
 

In [Fe(dipy)₃]²⁺ complex ion, Fe has 3d⁶ configuration. Due to pairing, this has no unpaired electrons.

Statement I [Ni(CN)₄]^2- is square planar and diamagnetic whereas [NiCI₄]^2- is tetrahedral and paramagnetic .
Statement II [Ni(CN)₄]^2- is square planar while [NiCI₄]^2- is tetrahedral.